How much work, in J, must be done on an ice bath to convert 8 g of ice at -5°C to steam at 10 0 ∘ C ? c ice = 0.5 cal / ( g ∘ C ) , L f = 80 , L v = 540 cal / g , J = 4.2 J / cal .

Heat = 8 [ 0.5 ( 5 ) + 80 + 100 + 540 ] = 8 ( 722.5 ) = 5780 cal . Work = 5780 × 4.2 = 24276 J .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermal ExpansionEasy2 minQB

PhysicsEasynumerical

How much work, in J, must be done on an ice bath to convert 8 g of ice at -5°C to steam at $10 0^{\circ} C ? c_{ice} = 0.5 cal / (g^{\circ} C), L_{f} = 80, L_{v} = 540 cal / g, J = 4.2 J / cal$ .

Answer:

24276.00

Solution:

Heat $= 8 [0.5 (5) + 80 + 100 + 540] = 8 (722.5) = 5780 cal$ . Work $= 5780 \times 4.2 = 24276 J$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

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